Christmas Streak 73/88: Geometry Rush #3

Geometry Level 3

In a parallelogram $\rm ABCD,$ draw a circle with diameter $\rm \overline{AB},$ and pick a point $\rm E$ on the circle such that it's on the interior of the parallelogram.

It is given that $\overrightarrow{\rm BE}$ meets with $\overline{\rm CD}$ at point $\rm F,$ where $\rm F$ is the midpoint of $\overline{\rm CD},$ and that $\rm C$ is on $\overrightarrow{\rm AE}.$

We also know that $\angle \rm FBC = 25^{\circ}.$ Find the size of $\angle \rm CED$ in degrees.

This problem is a part of <Christmas Streak 2017> series .

The answer is 115.0.

1 solution

Boi (보이)
Dec 18, 2017

Look at this thing on the right $\longrightarrow$

Aha, you notice that $\rm \overline{BC}=\overline{DP}$ by congruence of $\rm \triangle BCF \equiv \triangle PDF$

Now since $\rm D$ is the circumcenter of $\triangle \rm AEP,$ we figure that $\rm \overline{DP}=\overline{DE}.$

Note that $\rm \angle DPE=\angle CBE=25^{\circ}.$

We now know that $\rm \angle DEP=25^{\circ}.$

Since $\rm C$ is on $\overrightarrow{\rm AE},$ we already know $\angle \rm CEF=90^{\circ}.$

$\therefore~\angle \rm CFD=\boxed{115^{\circ}}$

Christmas Streak 73/88: Geometry Rush #3

The answer is 115.0.

1 solution

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